1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trig questions

  1. Jan 24, 2006 #1
    1. Express [tex] \frac{sin^2 2 \theta}{1+cos^2 2 \theta} [/tex] as a function of [tex]sin \theta[/tex]

    here's what I did:
    [tex] = \frac{4 sin^2 \theta cos^2 \theta}{1+(2cos^2 \theta -1)^2} [/tex]
    [tex] = \frac{4 sin^2 \theta cos^2 \theta}{1+(4cos^4 \theta - 4 cos^2 \theta + 1)} [/tex]
    [tex] = \frac{2sin^2 \theta cos^2 \theta}{2cos^2 \theta - 2cos^2 \theta +1} [/tex]

    is this correct? can I simplify it more?

    2. Write [tex]sin(a+b)sin(a-b)[/tex] as a function of double angles

    I used the sum and product formulas to simplify the equation but I did not use the double angle formulas. I'm not quite sure what the question is asking.

    Here's what I did:
    [tex] = .5 [cos(a+b-a+b)-cos(a+b+a-b)] [/tex]
    [tex] = .5 [cos(2b)-cos(2a)] [/tex]

    not sure where the double anges come in. any ideas?

    3. simplify [tex] \sqrt{2-2cos4 \Theta}[/tex]

    i can make it become 1-cos4x but I dont know how to simplify it further because of the cos4. no clue on this one, any help would be appreciated.
     
    Last edited: Jan 24, 2006
  2. jcsd
  3. Jan 24, 2006 #2

    TD

    User Avatar
    Homework Helper

    Well it's not written as a function of only sin(x), so it's not correct yet.
    Although there may be ways to simplify, you can reduce the sin²2x in function of only sinx like this:

    [tex]\sin ^2 2x = 4\sin ^2 x\cos ^2 x = 4\sin ^2 x\left( {1 - \sin ^2 x} \right)[/tex]

    In the denominator, there's a cos²2x but that's equal to 1-sin²2x, giving you the case above again.

    I'm not sure but perhaps this was the point, you have now rewritten it as a function of the double angles 2a and 2b.

    There are 3 version of the double-angle formule for the cosine, choose the one which makes the constant disappear:

    [tex]\sqrt {2 - 2\cos 4x} = \sqrt {2 - 2\left( {1 - 2\sin ^2 2x} \right)} [/tex]
     
  4. Jan 24, 2006 #3
    1.
    [tex]\frac {\sin^2 2 \theta} {2- \sin^2 2\theta} = \frac 2 {2 - \sin^2 2\theta} - 1[/tex]
    and simplify the denominator using [tex]\sin 2\theta = 2 \sin \theta \cos \theta[/tex] and [tex]\cos^2 \theta = 1 - \sin^2 \theta[/tex]...? Still complicated.

    2. looks correct.

    3. [tex] \sqrt{ 2( 1 - \cos 4\theta)} = \sqrt {2( 1 - \cos^2 2\theta + \sin^2 2\theta)} = 2 | \sin 2\theta} | = 4 | \sin \theta \cos \theta }|[/tex]

    hmmm...I have to go to see a class day of the kindergarten of my daughter next month... :)
     
    Last edited: Jan 24, 2006
  5. Jan 24, 2006 #4
    Your two problems can be solved easily-------------Akash

    Sin(a-b)Sin(a+b)
    =Sin(a^2) -Sin(b^2) /////////////////////you can check this easily. it is a formula
    =[1-cos(2a)-{1-Cos2b)]/2
    =[1-cos(2a)-1+cos(2b)]/2
    =[cos(2a)-cos(2b)]/2 ////////////////expressed in double angle in cos





    (2-2cos4A)^½=(2(1-cos4A))^½ ////////////A IS ANGLE THETA
    =(2(2Sin^2A))^½
    =(4Sin^2A)^½
    =2Sin^2A)------------------------------Soved

    Akash
    akash_413@sify.com
    captainvyom_akash@yahoo.com
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Trig questions
  1. Trig question (Replies: 4)

  2. Trig Question (Replies: 1)

  3. Trig question (Replies: 5)

  4. Trig question (Replies: 2)

  5. Trig question (Replies: 16)

Loading...